Solutions manual for introduction to genetic analysis 10th edition by griffiths wessler carroll and doebley

Answer: Yes, it is possible to find a pod with only yellow peas or heterozygous for the seed color gene, if all the flowers had dominant allele in a given fruit/pod.. In Figure 2-12, ass

B.Carroll and John Doebley

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Answer: No, the results would be different While self pollination produces 3 : 1 ratio

of yellow versus gene phenotype, cross pollination would result in 1 : 1 ratio, in the F2 This is because F1 yellow are heterozygous, while green are homozygous genotypes

2 In the right-hand part of Figure 2-4, in the plant showing an 11 : 11 ratio, do you think it would be possible to find a pod with all yellow peas? All green? Explain

Answer: Yes, it is possible to find a pod with only yellow peas or heterozygous for the seed color gene, if all the flowers had dominant allele in a given fruit/pod This could be also one example of rare changes at a physiological level

Answer: No, it should say either: “pairing, recombination, segregation, segregation” or: “replication, pairing, segregation, segregation.”

6 In Figure 2-12, assume (as in corn plants) that A encodes an allele that produces starch in pollen and allele a does not Iodine solution stains starch black How would you demonstrate Mendel’s first law directly with such a system?

Answer: One would use this iodine dye to color the starch producing corn pollen Since pollen is a plant gametophyte generation (haploid) it will be produced by meiosis Mendel’s first law predicts segregation of alleles into gametes, therefore we would expect 1 : 1 ratio of starch producing (A) versus non-starch producing (a) pollen grains, from a heterozygous (A/a) parent/male flower It would be easy to color the pollen and count the observed ratio

7 In the text figure on page 43, assume the left-hand individual is selfed What pattern of radioactive bands would you see in a Southern analysis of the progeny?

Answer: If an individual is selfed, the restriction fragments should be identical to the parents fragments In this case, a heterozygous parent to the left had three bands (two from a mutant allele “a” and one from dominant allele “A”)

8 Considering Figure 2-15, if you had a homozygous double mutant m3/m3 m5/m5,

would you expect it to be mutant in phenotype? (Note: This line would have two

mutant sites in the same coding sequence.)

9 In which of the stages of the Drosophila life cycle (represented in the box on page

52) does meiosis take place?

Answer: Meiosis happens in adult ovaries and testes, therefore before fertilization After fertilization, fruit flies would lay their eggs (with now diploid embryos) That would be Stage 1 on the figure

10 If you assume Figure 2-17 also applies to mice and you irradiate male sperm with

X rays (known to inactivate genes), what phenotype would you look for in

progeny in order to find cases of individuals with an inactivated SRY gene?

Answer: If we inactivate the SRY gene in mammals with radiation, the offspring should

all be phenotypically females, yet on the chromosome level there would be both

XX and XY (in this case sterile, female looking males)

11 In Figure 2-19, how does the 3 : 1 ratio in the bottom-left-hand grid differ from

the 3 : 1 ratios obtained by Mendel?

12 In Figure 2-21, assume that the pedigree is for mice, in which any chosen cross

can be made If you bred IV-1 with IV-3, what is the probability that the first baby will show the recessive phenotype?

Answer: The answer would be:

2/3 2/3 1/4 = 1/9 or 0.11

Probability that IV 1 and IV 3 mice are heterozygous is 2/3 This is because both of

their parents are known heterozygotes (A/a) and since they are dominant

phenotype they could only be A/A or A/a Now, probability that two heterozygotes have a recessive homozygote offspring is 1/4

13 Which part of the pedigree in Figure 2-23 in your opinion best demonstrates

Mendel’s first law?

Answer: Any part of this pedigree demonstrates the law, showing segregation of alleles into gametes The middle part of generation II marriage shows a typical test cross (expected 1:1) Neither ratio in the pedigree could be confirmed because of a small sample size in any given family, but allele segregation is obvious

Answer: Yes, it could in some cases, but in this case we have clues that the pedigree

is for a sex linked dominant trait First, if fathers have a gene, daughters will receive it only, and second, if mother has a gene, both sons and daughters would receive it

16 Peas (Pisum sativum) are diploid and 2n = 14 In Neurospora, the haploid fungus,

n = 7 If it were possible to fractionate genomic DNA from both species by using

pulsed field electrophoresis, how many distinct DNA bands would be visible in each species?

Answer: PFGE separates DNA molecules by size When DNA is carefully isolated

from Neurospora (which has seven different chromosomes) seven bands should

be produced using this technique Similarly, the pea has seven different chromosomes and will produce seven bands (homologous chromosomes will co-migrate as a single band)

17 The broad bean (Vicia faba) is diploid and 2n = 18 Each haploid chromosome

set contains approximately 4 m of DNA The average size of each chromosome during metaphase of mitosis is 13 m What is the average packing ratio of DNA

at metaphase? (Packing ratio = length of chromosome/length of DNA molecule therein.) How is this packing achieved?

Answer: There is a total of 4 m of DNA and nine chromosomes per haploid set On average, each is 4/9 m long At metaphase, their average length is 13 µm, so the

average packing ratio is 13 10–6

m : 4.4 10–1

m or roughly 1 : 34,000! This remarkable achievement is accomplished through the interaction of the DNA with proteins At its most basic, eukaryotic DNA is associated with histones in units called nucleosomes and during mitosis, coils into a solenoid As loops, it associates with and winds into a central core of nonhistone protein called the scaffold

18 If we call the amount of DNA per genome “x,” name a situation or situations in

diploid organisms in which the amount of DNA per cell is:

Answer: Because the DNA levels vary four-fold, the range covers cells that are haploid (gametes) to cells that are dividing (after DNA has replicated but prior to cell division) The following cells would fit the DNA measurements:

x + haploid cells

2x diploid cells in G1 or cells after meiosis I but prior to meiosis II

4x diploid cells after S but prior to cell division

21 Can you design a different nuclear-division system that would achieve the same

outcome as that of meiosis?

Answer: It’s pretty hard to beat several billions of years of evolution, but it might be simpler if DNA did not replicate prior to meiosis The same events responsible for halving the DNA and producing genetic diversity could be achieved in a single cell division if homologous chromosomes paired, recombined, randomly aligned

during metaphase, and separated during anaphase, etc However, you would lose the chance to check and repair DNA that replication allows

22 In a possible future scenario, male fertility drops to zero, but, luckily, scientists

develop a way for women to produce babies by virgin birth Meiocytes are converted directly (without undergoing meiosis) into zygotes, which implant in the usual way What would be the short- and long-term effects in such a society?

Answer: In large part, this question is asking, why sex? Parthenogenesis (the ability to reproduce without fertilization—in essence, cloning) is not common among multicellular organisms Parthenogenesis occurs in some species of lizards and fishes, and several kinds of insects, but it is the only means of reproduction in only a few of these species In plants, about 400 species can reproduce asexually

by a process called apomixis These plants produce seeds without fertilization However, the majority of plants and animals reproduce sexually Sexual reproduction produces a wide variety of different offspring by forming new combinations of traits inherited from both the father and the mother Despite the numerical advantages of asexual reproduction, most multicellular species that have adopted it as their only method of reproducing have become extinct However, there is no agreed upon explanation of why the loss of sexual reproduction usually leads to early extinction or conversely, why sexual reproduction is associated with evolutionary success

On the other hand, the immediate effects of such a scenario are obvious All offspring will be genetically identical to their mothers, and males would be extinct within one generation

“sister” chromatids are likely to be different Recombination during earlier meiotic stages has swapped regions of DNA between sister and nonsister chromosomes such that the two daughter cells of this division typically are not genetically identical

24 Make up mnemonics for remembering the five stages of prophase I of meiosis

and the four stages of mitosis

Answer: The four stages of mitosis are: prophase, metaphase, anaphase, and telophase The first letters, PMAT, can be remembered by a mnemonic such as: Playful Mice Analyze Twice

The five stages of prophase I are: leptotene, zygotene, pachytene, diplotene, and diakinesis The first letters, LZPDD, can be remembered by a mnemonic such as: Large Zoos Provide Dangerous Distractions

25 In an attempt to simplify meiosis for the benefit of students, mad scientists

develop a way of preventing premeiotic S phase and making do with having just one division, including pairing, crossing over, and segregation Would this system work, and would the products of such a system differ from those of the present system?

Answer: Yes, it could work but certain DNA repair mechanisms (such as postreplication recombination repair) could not be invoked prior to cell division There would be just two cells as products of this meiosis, rather than four

27 Francis Galton, a geneticist of the pre-Mendelian era, devised the principle that

half of our genetic makeup is derived from each parent, one-quarter from each grandparent, one-eighth from each great-grandparent, and so forth Was he right? Explain

28 If children obtain half their genes from one parent and half from the other parent,

why aren’t siblings identical?

Answer: Because the “half” inherited is very random, the chances of receiving exactly the same half is vanishingly small Ignoring recombination and focusing just on which chromosomes are inherited from one parent (for example, the one they inherited from their father or the one from their mother?), there are 223 = 8,388,608 possible combinations!

29 State where cells divide mitotically and where they divide meiotically in a fern, a

moss, a flowering plant, a pine tree, a mushroom, a frog, a butterfly, and a snail

gametophyte

sporophyte (anther and ovule)

pine tree sporophyte

gametophyte

sporophyte (pine cone)

mushroom sporophyte

gametophyte

sporophyte (ascus or basidium)

frog somatic cells gonads

butterfly somatic cells gonads

snail somatic cells gonads

30 Human cells normally have 46 chromosomes For each of the following stages,

state the number of nuclear DNA molecules present in a human cell:

Answer: This problem is tricky because the answers depend on how a cell is defined

In general, geneticists consider the transition from one cell to two cells to occur with the onset of anaphase in both mitosis and meiosis, even though cytoplasmic division occurs at a later stage

a 46 chromosomes, each with two chromatids = 92 chromatids

b 46 chromosomes, each with two chromatids = 92 chromatids

c 46 physically separate chromosomes in each of two about-to-be-formed cells

d 23 chromosomes in each of two about-to-be-formed cells, each with two

chromatids = 46 chromatids

e 23 chromosomes in each of two about-to-be-formed cells

31 Four of the following events are part of both meiosis and mitosis, but only one is

meiotic Which one? (1) chromatid formation, (2) spindle formation, (3) chromosome condensation, (4) chromosome movement to poles, (5) synapsis

Answer: (5) chromosome pairing (synapsis)

32 In corn, the allele ƒ´ causes floury endosperm and the allele f´´ causes flinty

endosperm In the cross ƒ´/ƒ´ [female symbol] ƒ´´/ƒ´´ [male symbol], all the

progeny endosperms are floury, but in the reciprocal cross, all the progeny endosperms are flinty What is a possible explanation? (Check the legend for Figure 2-7.)

34 If you had a fruit fly (Drosophila melanogaster) that was of phenotype A, what

test would you make to determine if the fly’s genotype was A/A or A/a?

35 In examining a large sample of yeast colonies on a petri dish, a geneticist finds an

abnormal-looking colony that is very small This small colony was crossed with wild type, and products of meiosis (ascospores) were spread on a plate to produce colonies In total, there were 188 wild-type (normal-size) colonies and 180 small ones

a What can be deduced from these results regarding the inheritance of the colony phenotype? (Invent genetic symbols.)

small-b What would an ascus from this cross look like?

Answer:

a A diploid meiocyte that is heterozygous for one gene (for example, s+/s where s is

the allele that confers the small colony phenotype) will, after replication and

segregation, give two meiotic products of genotype s+ and two of s If the random

spores of many meiocytes are analyzed, you would expect to find about 50 percent normally sized colonies and 50 percent small colonies if the abnormal phenotype is the result of a mutation in a single gene Thus, the actual results of 188 normally sized and 180 small-sized colonies support the hypothesis that the phenotype is the result of a mutation in a single gene

b The following represents an ascus with four spores The important detail is that two

of the spores are s and two are s+

36 Two black guinea pigs were mated and over several years produced 29 black and

9 white offspring Explain these results, giving the genotypes of parents and progeny

Answer: The progeny ratio is approximately 3:1, indicating classic

heterozygous-by-heterozygous mating Since black (B) is dominant to white (b): Parents: B/b B/b

Progeny: 3 black:1 white (1 B/B : 2 B/b : 1 b/b)

37 In a fungus with four ascospores, a mutant allele lys-5 causes the ascospores

bearing that allele to be white, whereas the wild-type allele lys-5+ results in black

ascospores (Ascospores are the spores that constitute the four products of meiosis.) Draw an ascus from each of the following crosses:

a You expect two lys-5+ (black) spores and two lys-5 (white) spores

c You expect all lys-5+ (black) spores

38 For a certain gene in a diploid organism, eight units of protein product are needed

for normal function Each wild-type allele produces five units

a If a mutation creates a null allele, do you think this allele will be recessive or mutant?

b What assumptions need to be made to answer part a?

Answer:

a You do not expect the mutation to be recessive This would be an example of a haploinsufficient gene since one copy of the wild-type allele does produce enough protein product for normal function

b An important assumption would be that having five of eight units of protein product would result in an observable phenotype It also assumes that the regulation of the single wild-type allele is not affected Finally, if the mutant allele was leaky rather than null, there might be sufficient protein function when heterozygous with a wild-type allele

39 A Neurospora colony at the edge of a plate seemed to be sparse (low density) in

comparison with the other colonies on the plate This colony was thought to be a possible mutant, and so it was removed and crossed with a wild type of the opposite mating type From this cross, 100 ascospore progeny were obtained None of the colonies from these ascospores was sparse, all appearing to be normal What is the simplest explanation of this result? How would you test your

explanation? (Note: Neurospora is haploid.)

Answer: The simplest explanation is that the abnormal phenotype was not due to an genetic change Perhaps the environment (edge of plate) was less favorable for

growth Since Neurospora is haploid and forms ascospores, isolating individual

asci from a cross of the possible “mutant” to wild type and individually growing the spores should yield 50 percent wild-type and 50 percent “mutant” colonies If all spores yield wild-type colonies, the low density phenotype was not heritable

40 From a large-scale screen of many plants of Collinsia grandiflora, a plant with

three cotyledons was discovered (normally, there are two cotyledons) This plant was crossed with a normal pure-breeding wild-type plant, and 600 seeds from this cross were planted There were 298 plants with two cotyledons and 302 with three cotyledons What can be deduced about the inheritance of three cotyledons? Invent gene symbols as part of your explanation

41 In the plant Arabidopsis thaliana, a geneticist is interested in the development of

trichomes (small projections) A large screen turns up two mutant plants

(A and B) that have no trichomes, and these mutants seem to be potentially useful

in studying trichome development (If they were determined by single-gene mutations, then finding the normal and abnormal functions of these genes would

be instructive.) Each plant is crossed with wild type; in both cases, the next generation (F1) had normal trichomes When F1 plants were selfed, the resulting

F2’s were as follows:

F2 from mutant A: 602 normal; 198 no trichomes

F2 from mutant B: 267 normal; 93 no trichomes

a What do these results show? Include proposed genotypes of all plants in your answer

b Under your explanation to part a, is it possible to confidently predict the F1 from crossing the original mutant A with the original mutant B?

Answer:

a The data for both crosses suggest that both A and B mutant plants are homozygous

for recessive alleles Both F2 crosses give 3:1 ratios of normal to mutant

progeny For example, let A = normal and a = mutant, then

P A / A a/a

F1 A/a

F2 1 A/A phenotype: normal

2 A/a phenotype: normal

1 a a phenotype: mutant (no trichomes)

b No You do not know if the a and b mutations are in the same or different genes If

they are in the same gene then the F1 will all be mutant If they are in different genes, then the F1 will all be wild type

42 You have three dice: one red (R), one green (G), and one blue (B) When all three

dice are rolled at the same time, calculate the probability of the following outcomes:

independent So each independent probability is multiplied together a

e The easiest way to approach this problem is to consider each die separately

The first die thrown can be any number Therefore, the probability for it is 1 The second die can be any number except the number obtained on the first die

Therefore, the probability of not duplicating the first die is 1 – p(first die

duplicated) = 1 – 1/6 = 5/6 The third die can be any number except the numbers obtained on the first two

dice Therefore, the probability is 1 – p(first two dice duplicated) = 1 – 2/6 =

2

/3 Finally, the probability of all different dice is (1)(5/6)(2/3) = 10/18 = 5/9

44 a The ability to taste the chemical phenylthiocarbamide is an autosomal dominant

phenotype, and the inability to taste it is recessive If a taster woman with a nontaster father marries a taster man who in a previous marriage had a nontaster daughter, what is the probability that their first child will be:

Answer:

a By considering the pedigree (see below), you will discover that the cross in question

is T/t T/t Therefore, the probability of being a taster is 3/4, and the probability of being a nontaster is 1/4

Also, the probability of having a boy equals the probability of having a girl equals 1/2

(1) p(nontaster girl) = p(nontaster) p(girl) = 1/4 1/2 = 1/8

(2) p(taster girl) = p(taster) p(girl) = 3/4 1/2 = 3/8

(3) p(taster boy) = p(taster) p(boy) = 3/4 1/2 = 3/8

b p(taster for first two children) = p(taster for first child) p(taster for second child)

= 3/4 3/4 = 9/16

45 John and Martha are contemplating having children, but John’s brother has

galactosemia (an autosomal recessive disease) and Martha’s great-grandmother also had galactosemia Martha has a sister who has three children, none of whom have galactosemia What is the probability that John and Martha’s first child will have galactosemia?

?

genotypes can be stated only as G/– for both John and Martha

The probability that John is carrying the allele for galactosemia is 2/3, rather than the 1/2 that you might guess To understand this, recall that John’s parents must be heterozygous in order to have a child with the recessive disorder while still being normal themselves (the assumption of normalcy is based

on the information given in the problem) John’s parents were both G/g A

Punnett square for their mating would be:

The expected ratio of the F1 is 1 G/G : 2 G/g : 1 g/g Because John does not have

galactosemia (an assumption based on the information given in the

problem), he can be either G/G or G/g, which occurs at a ratio of 1:2 Therefore, his probability of carrying the g allele is 2/3

The probability that Martha is carrying the g allele is based on the following

chain of logic Her great-grandmother had galactosemia, which means that she had to pass the allele to Martha’s grandparent Because the problem states nothing with regard to the grandparent’s phenotype, it must be

assumed that the grandparent was normal, or G/g The probability that the

grandparent passed it to Martha’s parent is 1

/2 Next, the probability that Martha’s parent passed the allele to Martha is also 1

/2, assuming that the parent actually has it Therefore, the probability that Martha’s parent has the allele and passed it to Martha is 1/2 1/2, or 1/4

/4

This information does not fit easily into a Punnett square

Galactosemia is a metabolic disorder characterized by the absence of the enzyme

galactose-1-phosphate uridyl transferase, which results in an accumulation

of galactose In the vast majority of cases, galactosemia results in an enlarged liver, jaundice, vomiting, anorexia, lethargy, and very early death

if galactose is not omitted from the diet (initially, the child obtains galactose from milk)

Autosomal refers to genes that are on the autosomes

Recessive means that in order for an allele to be expressed, it must be the only

form of the gene present in the organism

Answer: Autosomal recessive disorders are assumed to be rare and to occur equally frequently in males and females They are also assumed to be expressed if the person is homozygous for the recessive genotype

10 What is the relevance of the rareness of the phenotype under study in pedigree analysis generally, and what can be inferred in this problem?

Answer: Rareness leads to the assumption that people who marry into a family that is being studied do not carry the allele, which was assumed in entry (6) above

13 Is there any irrelevant information in the problem as stated?

Answer: The information regarding Martha’s sister and her children turns out to

be irrelevant to the problem

15 Can you make up a short story based on the human dilemma in this problem?

46 Holstein cattle are normally black and white A superb black-and-white bull,

Charlie, was purchased by a farmer for $100,000 All the progeny sired by Charlie were normal in appearance However, certain pairs of his progeny, when interbred, produced red-and-white progeny at a frequency of about 25 percent Charlie was soon removed from the stud lists of the Holstein breeders Use symbols to explain precisely why

Answer: Charlie, his mate, or both, obviously were not homozygous for one of the alleles (pure-breeding), because his F2 progeny were of two phenotypes Let A = black and white, and a = red and white If both parents were heterozygous, then

red and white would have been expected in the F1 generation Red and white were not observed in the F1 generation, so only one of the parents was heterozygous The cross is:

However, if the F2 progeny came only from one mate, the farmer may have acted too quickly

47 Suppose that a husband and wife are both heterozygous for a recessive allele for

albinism If they have dizygotic (two-egg) twins, what is the probability that both the twins will have the same phenotype for pigmentation?

Answer: Because the parents are heterozygous, both are A/a Both twins could be

albino or both twins could be normal (and = multiply, or = add) The probability

of being normal (A/–) is 3/4, and the probability of being albino (a/a) is 1/4

p(both normal) + p(both albino)

p(first normal) p(second normal) + p(first albino) p(second albino)

(3/4)(3/4) + (1/4)(1/4) = 9/16 + 1/16 = 5/8

48 The plant blue-eyed Mary grows on Vancouver Island and on the lower mainland

of British Columbia The populations are dimorphic for purple blotches on the leaves—some plants have blotches and others don’t Near Nanaimo, one plant in nature had blotched leaves This plant, which had not yet flowered, was dug up and taken to a laboratory, where it was allowed to self Seeds were collected and grown into progeny One randomly selected (but typical) leaf from each of the progeny is shown in the accompanying illustration

a Formulate a concise genetic hypothesis to explain these results Explain all

symbols and show all genotypic classes (and the genotype of the original plant)

b How would you test your hypothesis? Be specific

Answer: The plants are approximately 3 blotched : 1 unblotched This suggests that blotched is dominant to unblotched and that the original plant which was selfed was a heterozygote

a Let A = blotched, a = unblotched

P A/a (blotched) A/a (blotched)

F1 1 A/A : 2 A/a : 1 a/a

3 A/– (blotched) : 1 a/a (unblotched)

b All unblotched plants should be pure-breeding in a testcross with an

unblotched plant (a/a), and one-third of the blotched plants should be

pure-breeding

49 Can it ever be proved that an animal is not a carrier of a recessive allele (that is,

not a heterozygote for a given gene)? Explain

Answer: In theory, it cannot be proved that an animal is not a carrier for a recessive

allele However, in an A/– a/a cross, the more dominant-phenotype progeny produced, the less likely it is that the parent is A/a In such a cross, half the progeny would be a/a and half would be A/a With n dominant phenotype progeny, the probability that the parent is A/a is (1/2)n (DNA sequencing can be used to prove heterozygosity, but without sequence level information, the level of certainty is limited by sample size.)

50 In nature, the plant Plectritis congesta is dimorphic for fruit shape; that is,

individual plants bear either wingless or winged fruits, as shown in the illustration Plants were collected from nature before flowering and were crossed

or selfed with the following results:

Answer: The results suggest that winged (A/–) is dominant to wingless (a/a) (cross 2

gives a 3 : 1 ratio) If that is correct, the crosses become:

The five unusual plants are most likely due either to human error in classification or to contamination Alternatively, they could result from environmental effects on development For example, too little water may have prevented the seed pods from becoming winged even though they are genetically winged

a How is the disorder inherited? State reasons for your answer

b Give genotypes for as many individuals in the pedigree as possible (Invent your own defined allele symbols.)

c Consider the four unaffected children of parents III-4 and III-5 In all four-child progenies from parents of these genotypes, what proportion is expected to contain all unaffected children?

Answer:

a The disorder appears to be dominant because all affected individuals have an affected parent If the trait was recessive, then I-1, II-2, III-1, and III-8 would all have to be carriers (heterozygous for the rare allele)

c The mating is D/d d/d The probability of an affected child (D/d) equals 1/2, and

the probability of an unaffected child (d/d) equals 1/2 Therefore, the chance of having four unaffected children (since each is an independent event) is: 1/2 1/2 1

/2 1/2 = 1/16

52 Four human pedigrees are shown in the accompanying illustration The black

symbols represent an abnormal phenotype inherited in a simple Mendelian manner